Expression For Work Done During isothermal Process - Thermodynamics 

Consider one mole of an ideal gas contained in a perfectly conducting cylinder fitted with a perfectly frictionless piston of area of cross section area A.

Let the volume of the gas be V exerting a pressure P on the walls of the cylinder and the piston. Let the gas expands and the piston moves through a small distance dx.

Force acting on the piston , F = PA

$\therefore$ work done to move the piston through a distance dx is given by

$dW = Fdx= PA dx = PdV$

To find the total work done from volume $( V_1) to ( V_2)$:

$W = \int_{V_1}^{V_2} P \, dV$

Step 2: Use the Ideal Gas Law

For an ideal gas:

$PV = nRT \Rightarrow P = \frac{nRT}{V}$

Substituting into the work integral:

$W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV$

Step 3: Take Constants Outside the Integral

Since \( n \), \( R \), and \( T \) are constants during an isothermal process:

$W = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV$

Step 4: Integrate

$\int_{V_1}^{V_2} \frac{1}{V} \, dV = \ln\left(\frac{V_2}{V_1}\right)$

$\Rightarrow W = nRT \ln\left(\frac{V_2}{V_1}\right)$

Final Expression (Natural Log Form)

$\boxed{W = nRT \ln\left(\frac{V_2}{V_1}\right)}$

Step 5: Convert to Base-10 Logarithm

Using the identity $( \ln x = 2.303 \log_{10} x)$:

$W = 2.303 \cdot nRT \cdot \log_{10}\left(\frac{V_2}{V_1}\right)$

Final Expression (Base-10 Log Form)

$\boxed{W = 2.303 \cdot nRT \cdot \log_{10}\left(\frac{V_2}{V_1}\right)}$

Step 6: Express in Terms of Pressure

From Boyle's Law:

$P_1 V_1 = P_2 V_2 \Rightarrow \frac{V_2}{V_1} = \frac{P_1}{P_2}$

Substitute into the expression:

$W = 2.303 \cdot nRT \cdot \log_{10}\left(\frac{P_1}{P_2}\right)$

Final Expression (Pressure-Based Form)}

$\boxed{W = 2.303 \cdot nRT \cdot \log_{10}\left(\frac{P_1}{P_2}\right)}$

Step 7: First Law of Thermodynamics

From the first law:

$Q=\Delta U+ W$

In an isothermal process, $( \Delta U = 0)$, so:

$Q = W$

Final Heat Transfer Relation

$\boxed{Q = W = 2.303 \cdot nRT \cdot \log_{10}\left(\frac{V_2}{V_1}\right)}$

Sign Convention

If $( V_2 > V_1)$: Expansion, $( W > 0), ( Q > 0 )$ (heat absorbed)

If $( V_2 < V_1)$: Compression, $( W < 0), ( Q < 0)$ (heat released).